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F(y)=summation 1/(y^2+m^2) is not differentiable.

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Homework Statement



Is [tex] f(y)=\sum_{m=1}^\infty \frac{1}{y^2+m^2}
[/tex]differentiable?

Homework Equations





The Attempt at a Solution



From the graph, it is obvious that f is not differentiable at y=0, but I don't know how to prove that. I proved that [tex] \sum_{m=1}^n f_m=\sum_{m=1}^n\frac{1}{y^2+m^2} [/tex] converges uniformly to f. Does this help? Thank you very much.
 

Answers and Replies

  • #2
Dick
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The uniform convergence of differentiable functions does not prove the sum is differentiable. If the derivatives converge uniformly then it is. Don't you have a theorem like that? And why is it clear it's not differentiable at zero?
 
  • #3
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The uniform convergence of differentiable functions does not prove the sum is differentiable. If the derivatives converge uniformly then it is. Don't you have a theorem like that? And why is it clear it's not differentiable at zero?
I have been thinking that each [tex] f_m[/tex] is strictly increasing on [tex](0,\infty)[/tex]. Therefore, so is f(y). If f is differentiable, then f'(0)=0 which is not true. Is this correct?
 
  • #4
Dick
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I have been thinking that each [tex] f_m[/tex] is strictly increasing on [tex](0,\infty)[/tex]. Therefore, so is f(y). If f is differentiable, then f'(0)=0 which is not true. Is this correct?
Do you mean decreasing? f'(0) has to be zero, the function is even. 1/(x^2+1) is strictly decreasing on [0,infinity). It's perfectly differentiable.
 
  • #5
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Do you mean decreasing? f'(0) has to be zero, the function is even. 1/(x^2+1) is strictly decreasing on [0,infinity). It's perfectly differentiable.
Yeah I meant decreasing.
I know that each [tex]f_m[/tex] is differentiable on R. But f is not. I don't know how to prove this though. I am trying to prove by contradiction, but I cannot find a contradiction.
 
  • #6
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Do you mean decreasing? f'(0) has to be zero, the function is even. 1/(x^2+1) is strictly decreasing on [0,infinity). It's perfectly differentiable.
I think that f is not differentiable at y=0 because the graph of f is really pointy at y=0...
 
  • #7
Dick
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I think that f is not differentiable at y=0 because the graph of f is really pointy at y=0...
If f(y) looks really pointy at y=0 it might be a problem with the graphing program. This is an analysis course, right? You shouldn't rely on that.
 
  • #8
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If f(y) looks really pointy at y=0 it might be a problem with the graphing program. This is an analysis course, right? You shouldn't rely on that.
Hi,

Now I see how to show that it is differentiable at x=0, but still don't know how to show that it is differentiable on the rest of R. I have two tools: cauchy criterion and weierstrass m-test. I tried both but none worked. Can you give some hints? Thanks.
 
  • #9
Dick
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Hi,

Now I see how to show that it is differentiable at x=0, but still don't know how to show that it is differentiable on the rest of R. I have two tools: cauchy criterion and weierstrass m-test. I tried both but none worked. Can you give some hints? Thanks.
Like I said before, you want to prove the sum of the derivatives converges uniformly to f'(x). The weierstrass m-test should work fine. Can you show use where you are having trouble with it?
 
  • #10
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Like I said before, you want to prove the sum of the derivatives converges uniformly to f'(x). The weierstrass m-test should work fine. Can you show use where you are having trouble with it?
Hey i actually figured it out! I did not think about using fundamental calc to find a bound at first...Thanks! :)
 

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